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JAMB: Mathematics,
Question 28, 2013
Find the length of a chord which subtends an angle of 90 o at
the centre of a circle whose radius is 8cm (a) 8cm (b) 8 𝟐𝒄𝒎
(c) (d) 4cm
SOLUTION
Topic/Subtopic: Circles/Chord Properties
To solve this problem, let us show it on a diagram
Where 𝐴𝐵 = Length of the chord and the radius (r) = 8cm, cut out the ΔAOB
8cm8cm
0
A B
90o
Circle
Radius
Chord
Length 𝐴𝐵 can be obtained by cosine – Rule since two sides of a triangle and
an included angle are given (condition for applying cosine rule).
A B
0
90
8cm 8cm
Applying cosine – rule to the triangle we have:
• /AB/2 = /AO/2 + /BO/2 – 2/AO//BO/Cos 𝑂
• /AB/2 = 82 + 82 – (2 x 8 x 8 x cos 90o)
• /AB/2 = 64 + 64 – (128 x cos 90o)
• /AB/2 = 128 – (128 x 0) since cos 90o = 0
• /AB/2 = 128 – 0 = 128
• /AB/ = 128 = 2 𝑥 64 = 2 x 64 = 2 x 8
• Therefore, /AB/ = 8 𝟐𝒄𝒎 (Option B)
It is important to note that the triangle AOB is a right-angle triangle since angle
𝑂 = 90o as a result. We can also apply Pythagoras theorem to solve for /AB/ as
shown in the next slide (method 2):
METHOD II
We can as well draw the triangle like this:
A B
0
8cm 8cm
Applying Pythagoras theorem to the triangle we have:
/AB/2 = /AO/2 + /BO/2 (the square of the longest side equals the sum of the
square of the two other sides)
• /AB/2 = 82 + 82
• /AB/2 = 64 + 64
• /AB/2 = 128
• /AB/2 = 128
• /AB/2 = 2 x 64
• /AB/2 = 2 x 64
• /AB/2 = 2 x 8
• /AB/2 = 8 2𝑐𝑚
Final notes
(1) Pythagoras theorem is applied only when the triangle is a right-angled
triangle
(2) When two sides of a triangle are given and you are asked to find the third
side

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Jamb mathematics-question 28- 2013

  • 1. JAMB: Mathematics, Question 28, 2013 Find the length of a chord which subtends an angle of 90 o at the centre of a circle whose radius is 8cm (a) 8cm (b) 8 𝟐𝒄𝒎 (c) (d) 4cm
  • 2. SOLUTION Topic/Subtopic: Circles/Chord Properties To solve this problem, let us show it on a diagram Where 𝐴𝐵 = Length of the chord and the radius (r) = 8cm, cut out the ΔAOB 8cm8cm 0 A B 90o Circle Radius Chord
  • 3. Length 𝐴𝐵 can be obtained by cosine – Rule since two sides of a triangle and an included angle are given (condition for applying cosine rule). A B 0 90 8cm 8cm
  • 4. Applying cosine – rule to the triangle we have: • /AB/2 = /AO/2 + /BO/2 – 2/AO//BO/Cos 𝑂 • /AB/2 = 82 + 82 – (2 x 8 x 8 x cos 90o) • /AB/2 = 64 + 64 – (128 x cos 90o) • /AB/2 = 128 – (128 x 0) since cos 90o = 0 • /AB/2 = 128 – 0 = 128 • /AB/ = 128 = 2 𝑥 64 = 2 x 64 = 2 x 8 • Therefore, /AB/ = 8 𝟐𝒄𝒎 (Option B)
  • 5. It is important to note that the triangle AOB is a right-angle triangle since angle 𝑂 = 90o as a result. We can also apply Pythagoras theorem to solve for /AB/ as shown in the next slide (method 2):
  • 6. METHOD II We can as well draw the triangle like this: A B 0 8cm 8cm
  • 7. Applying Pythagoras theorem to the triangle we have: /AB/2 = /AO/2 + /BO/2 (the square of the longest side equals the sum of the square of the two other sides) • /AB/2 = 82 + 82 • /AB/2 = 64 + 64 • /AB/2 = 128 • /AB/2 = 128 • /AB/2 = 2 x 64 • /AB/2 = 2 x 64 • /AB/2 = 2 x 8 • /AB/2 = 8 2𝑐𝑚
  • 8. Final notes (1) Pythagoras theorem is applied only when the triangle is a right-angled triangle (2) When two sides of a triangle are given and you are asked to find the third side